pada kubus ABCD.EFGH, titik P,Q berturut-turut titik tengah rusuk GH dan FG. Jika alfa sudut antara bidang BDE dan BDPQ, maka sin alfa=

[tex]rusuk = x \\
rusuk \: diagonal \: sisi = x \sqrt{2} \\
rusuk \: diagonal \: ruang = x \sqrt{3} \\
O = titik \: tengah BD \\
N = titik \: tengah PQ \\ \\

\alpha = sudut (EBN) \\ \\

NE = \frac{3}{4} x \sqrt{2} \\

EO = \sqrt{{( \frac{1}{2} x)}^{2} + {( \frac{1}{2} x)}^{2} + {x}^{2}} \\ = \sqrt{ \frac{3}{2} {x}^{2} } = \frac{1}{2} x \sqrt{6} \\ \\ ON = \sqrt{( \frac{1}{4}x) ^{2} + {( \frac{1}{4}x) }^{2} + {x}^{2} } \\ = \frac{x}{4} \sqrt{20} = \frac{1}{2} x \sqrt{5} \\ \\ F = titik \: tengah \: ON\\ (EO) ^{2} - (O F) ^{2} = (NE) ^{2} - (ON - O F) ^{2} \\ {(\frac{1}{2} x \sqrt{6})}^{2} - (O F) ^{2} = {(\frac{3}{4} x \sqrt{2})}^{2} - (\frac{1}{2} x \sqrt{5} - (O F)) ^{2} \\ 3 {x}^{2} - (O F) ^{2} = \frac{9}{8} {x}^{2} - \frac{5}{4} {x}^{2} + x \sqrt{5} (O F) - {(O F)}^{2} \\ 3 \frac{1}{8} {x}^{2} = x \sqrt{5} (O F) \\ (O F) = \frac{25}{8} {x}^{2} \times \frac{ \sqrt{5} }{5x} = \frac{5}{8}x \sqrt{5} \\ \\ EF = \sqrt{(EO) ^{2} - (O F) ^{2}} \\ = \sqrt{{(\frac{1}{2} x \sqrt{6})}^{2} - {(\frac{5}{8}x \sqrt{5})}^{2} } \\ = \sqrt{3 {x}^{2} - \frac{125}{64} {x}^{2} } = \sqrt{ \frac{192 {x}^{2} - 125{x}^{2} }{64} } = \frac{x}{8} \sqrt{67} \\ \\ \sin( \alpha ) = \frac{EF}{EO} = \frac{ \sqrt{6}x \times 8 }{2 \times x \sqrt{67} } = 4 \sqrt{402} [/tex]